abode_x: you got
[tex] {i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]
=\delta\omega^\mu{}_\nu P^\nu [/tex]
Next,
[tex]\delta\omega^\mu{}_\nu P^\nu
=\delta\omega_{\rho\nu}g^{\rho\mu} P^\nu
=\delta\omega_{\rho\sigma}g^{\rho\mu} P^\sigma[/tex]
Next, use that [tex]\delta\omega_{\rho\sigma}[/tex] is antisymmetric, and so we can replace [tex] g^{\rho\mu} P^\sigma [/tex] with its antisymmetric part,
[tex]{1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]
This gives
[tex]{i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]={1\over2}\delta\omega_{\rho\sigma}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]
For this to be true for any [tex]\delta\omega_{\rho\sigma}[/tex], it must be true of each coefficient. So,
[tex]{i\over2\hbar}[P^\mu,M^{\rho\sigma}]
={1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]
Multiply by [tex]-2i\hbar[/tex] and switch the two terms on the right to get
[tex] [P^\mu,M^{\rho\sigma}] =i\hbar(g^{\sigma\mu}P^\rho-g^{\rho\mu} P^\sigma)[/tex]
Using [tex]g^{\sigma\mu}=g^{\mu\sigma}[/tex], this matches Srednicki's eq.(2.18).